8—Multivariable Calculus 182Iff(x,y) =x^2 +y^2 this better be zero, because I’m finding howfchanges whenris held fixed. Check
it out; it is. The equation (8.6) presents the form that is most important in many applications.
Example: What is the derivative ofywith respect toφat constantx?
(
∂y
∂φ
)
x=
(
∂y
∂r
)
φ(
∂r
∂φ
)
x+
(
∂y
∂φ
)
r(
∂φ
∂φ
)
x= [sinφ].
[
r
sinφ
cosφ
]
+ [rcosφ].1 =r
1
cosφ
(8.7)
φ
∆φ
r∆φ ∆y
You see a graphical interpretation of the calculation in this diagram: φchanges by∆φ, so the
coordinate moves up by∆y(xis constant). The angle between the lines∆yandr∆φisφitself. This
means that∆y÷r∆φ= 1/cosφ, and that is precisely the preceding equation for
(
∂y/∂φ
)
x.
In doing the calculation leading to Eq. (8.7), do you see how to do the calculation for(
∂r/∂φ
)
x?Differentiate the equationx=rcosφwith respect toφ.
x=rcosφ →
(
∂x
∂φ
)
x= 0 =
(
∂r
∂φ
)
xcosφ+r
(
∂cosφ
∂φ
)
x=
(
∂r
∂φ
)
xcosφ−rsinφ
Solve for the unknown derivative and you have the result.
Another example: f(x,y) = x^2 − 2 xy. The transformation between rectangular and polar
coordinates isx=rcosφ, y=rsinφ. What is
(
∂f/∂x
)
r?
(∂f
∂x
)
r=
(
∂f
∂x
)
y(
∂x
∂x
)
r+
(
∂f
∂y
)
x(
∂y
∂x
)
r= (2x− 2 y) + (− 2 x)
(
∂y
∂x
)
r
(∂y
∂x
)
r=
(
∂y/∂φ
)
( r∂x/∂φ
)
r=
rcosφ
−rsinφ
=−cotφ (8.8)
(Remember problem1.49?) Put these together and
(
∂f
∂x
)
r= (2x− 2 y) + (− 2 x)(−cotφ) = 2x− 2 y+ 2xcotφ (8.9)
The brute-force way to do this is to express the functionfexplicitly in terms of the variablesxandr,
eliminatingyandφ.
y=rsinφ=
√
( r^2 −x^2 , then
∂f
∂x
)
r=
∂
∂x
[
x^2 − 2 x
√
r^2 −x^2
]
r= 2x− 2
√
r^2 −x^2 − 2 x
1
√
r^2 −x^2
(−x) = 2x+
− 2
(
r^2 −x^2
)
+ 2x^2
√
r^2 −x^2
(8.10)
You can see that this is the same as the equation (8.9) if you look at the next-to-last form of equation
(8.10).